f(x)=2cos(x+π/4)cos(x-π/4)+√3sin2x的值域和最小正周期要过程,谢谢
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![f(x)=2cos(x+π/4)cos(x-π/4)+√3sin2x的值域和最小正周期要过程,谢谢](/uploads/image/z/3857947-43-7.jpg?t=f%28x%29%3D2cos%28x%2B%CF%80%2F4%29cos%28x-%CF%80%2F4%29%2B%E2%88%9A3sin2x%E7%9A%84%E5%80%BC%E5%9F%9F%E5%92%8C%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E8%A6%81%E8%BF%87%E7%A8%8B%2C%E8%B0%A2%E8%B0%A2)
f(x)=2cos(x+π/4)cos(x-π/4)+√3sin2x的值域和最小正周期要过程,谢谢
f(x)=2cos(x+π/4)cos(x-π/4)+√3sin2x的值域和最小正周期
要过程,谢谢
f(x)=2cos(x+π/4)cos(x-π/4)+√3sin2x的值域和最小正周期要过程,谢谢
有个公式:cos(a)cos(b)=1/2[cos(a+b)+cos(a-b)]
f(x)=2cos(x+π/4)cos(x-π/4)+√3sin2x
=cos(2x)+cos(π/2)+√3sin2x
=cos(2x)+√3sin2x
=2[cos(π/3)cos(2x)+sin(π/3)sin(2x)]
=2cos(2x-π/3)
值域:[-2,2]
最小正周期:π
楼主,多给些赏金呀~